JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let a = f'(1), b = f''(2), c = f'''(3). Then f(x) = x^3 + ax^2 + 2bx + c. Taking derivatives: f'(x) = 3x^2 + 2ax + 2b, f''(x) = 6x + 2a, f'''(x) = 6. From f'(1) = a: 3 + 2a + 2b = a gives a = -3 - 2b. From f''(2) = b: 12 + 2a = b. From f'''(3) = c: c = 6. Substituting the first into the second: b = 12 + 2(-3 - 2b) = 6 - 4b gives 5b = 6, so b = 6 5 . Then a = -3 - 12 5 = - 27 5 . f(x) = x^3 - 27 5 x^2 + 12 5 x + 6 f'(x) = 3x^2 - 54 5 x + 12 5 f'(5) = 75 - 54 + 12 5 = 117 5